Electricity

25

Reason — Given,

Wire of resistance R is cut into five equal parts hence, resistance of each part is $\dfrac{\text{R}}{5}$

$\dfrac{1}{\text{R}'} = \dfrac{1}{\dfrac{\text{R}}{5}} + \dfrac{1}{\dfrac{\text{R}}{5}}+ \dfrac{1}{\dfrac{\text{R}}{5}} + \dfrac{1}{\dfrac{\text{R}}{5}} + \dfrac{1}{\dfrac{\text{R}}{5}} \\[1em] = \dfrac{5}{\text{R}} + \dfrac{5}{\text{R}} + \dfrac{5}{\text{R}} + \dfrac{5}{\text{R}} + \dfrac{5}{\text{R}} \\[1em] = \dfrac{25}{\text{R}} \\[1em] \Rightarrow \text{R}' = \dfrac{\text{R}}{25}$

Hence,

$\dfrac{\text{R}}{\text{R}'} = \dfrac{\text{R}}{\dfrac{\text{R}}{25}} \\[1em] \Rightarrow \dfrac{\text{R}}{\text{R}'} = 25$

IR²

Reason — Electrical power in a circuit:

P = VI = I²R = $\dfrac{\text{V}^2}{\text{R}}$

25 W

Reason —

Given, V = 220V ; P = 100 W ; R = ?

We know,

$\phantom{\Rightarrow} \text{P} = \dfrac{\text{V}^2}{\text{R}} \\[1em] \Rightarrow \text{R} = \dfrac{\text{V}^2}{\text{P}}$

Substituting the values, we get

R = $\dfrac{\text{220}^2}{\text{100}}$ = 484 Ω

When voltage is reduced, resistance remains the same. Hence, the power consumed :

P = $\dfrac{110^2}{484}$ = 25 W

Hence, the power consumed = 25 W.

1:4

Reason — Let R_s and R_p be the equivalent resistance of the wires when connected in series and parallel respectively.

R_s = R + R = 2R

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{\text{R}} + \dfrac{1}{\text{R}} = \dfrac{2}{\text{R}}$

R_p = $\dfrac{\text{R}}{2}$

Let P_s and P_p be the power consumed in series and parallel circuits, respectively.

Power (P) = $\dfrac{\text{V}^2}{\text{R}}$

P_s = $\dfrac{\text{V}^2}{\text{2R}}$

and

P_p = $\dfrac{\text{V}^2}{\dfrac{\text{R}}{2}} = \dfrac{2\text{V}^2}{\text{R}}$

For the same potential difference V, the ratio of the heat produced in the circuit is given by ratio of P_s and P_p hence

$\text{H}_s : \text{H}_p = \dfrac{\dfrac{\text{V}^2}{\text{2R}}}{\dfrac{2\text{V}^2}{\text{R}}} \\[1em] = \dfrac{\text{V}^2 \times \text{R}}{\text{2R} \times 2\text{V}^2} \\[1em] = \dfrac{1}{4}$

Hence, the ratio of the heat produced is 1:4.

In the circuit, voltmeter is connected in parallel to measure the potential difference between two points.

Given,

diameter = 0.5 mm

area = $\dfrac{πd^2}{4} = \dfrac{3.14 \times (5 \times 10^{-4})^2}{4}$ = 19.625 x 10^–8 m

resistivity = 1.6 × 10^–8 Ω m

R = 10 Ω

l = ?

We know, R = ρ $\dfrac{l}{A}$

Substituting we get,

10 = 1.6 × 10^–8 x $\dfrac{l}{19.625 \times 10^{-8}}$

l = $\dfrac{10 \times 19.625}{1.6 }$ = 122.7 m

The length of the wire is 122.72 m.

When the diameter is doubled

According to the formula,

R = ρ $\dfrac{l}{A}$ where A = $\dfrac{πd^2}{4}$

Hence, R ∝ $\dfrac{1}{d^2}$

When diameter is double,

Resistance will change by $\dfrac{1}{(2d)^2}$ = $\dfrac{1}{4d^2}$

∴ Resistance will become $\dfrac{1}{4}^{\text{th}}$ of its original value.

Hence, new resistance = $\dfrac{1}{4} \times 10$ = 2.5 Ω

The graph is plotted as below:

The slope of the line indicates the value of resistance.

Let P₁(3.4, 1) and P₂(10.2, 3) be two points on the graph

Slope = $\dfrac{1}{\text{R}} = \dfrac{3-1}{10.2-3.4} = \dfrac{2}{6.8}$

R = $\dfrac{6.8}{2}$ = 3.4 Ω

Therefore, the resistance = 3.4 Ω.

Given,

V = 12 V

I = 2.5 mA = 2.5 x 10^-3A

R = ?

According to Ohm's Law:

V = IR

$\therefore \text{R} = \dfrac{\text{V}}{\text{I}} \\[1em] \Rightarrow \text{R} = \dfrac{12}{2.5 \times 10^{-3}} \\[1em] \Rightarrow \text{R} = 4.8 \times 10^{3} \Omega$

Hence, resistance of the resistor = 4.8 kΩ

Given,

V = 9V

Resistors in series = 0.2 Ω, 0.3 Ω, 0.4 Ω, 0.5 Ω and 12 Ω,

R_s = 0.2 + 0.3 + 0.4 + 0.5 + 12 = 13.4 Ω

According to Ohm's Law:

V = IR

$\therefore \text{I} = \dfrac{\text{V}}{\text{R}} \\[1em] \Rightarrow \text{I} = \dfrac{9}{13.4} \\[1em] \Rightarrow \text{I} = 0.67 \text{ A}$

Therefore, current flowing = 0.67 A.

Given,

I = 5 A

V = 220 V

Let number of resistors be n

Resistance of each resistor = 176 Ω

$\dfrac{1}{\text{R}_{eq}} = \dfrac{1}{176} + \dfrac{1}{176} + ....n \\[1em] \dfrac{1}{\text{R}_{eq}} = \dfrac{\text{n}}{176} \Omega \\[1em] \Rightarrow \text{R}_{eq} = \dfrac{176}{\text{n}} \Omega$

Applying Ohm's Law:

V = IR

220 = 5 x $\dfrac{176}{\text{n}}$

n = $\dfrac{176 \times 5}{220}$ = 4 resistors

Hence, 4 resistors are connected in parallel.

If we connect resistors in series, R = 6 Ω + 6 Ω + 6 Ω = 18 Ω.

If we connect all resistors in parallel,

$\dfrac{1}{\text{R}} = \dfrac{1}{6} + \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{3}{6} = \dfrac{1}{2}$

Hence, R = 2Ω

We can obtain the desired value by connecting two of the resistors in either series or parallel.

Case (i)

If two resistors are connected in parallel, then their equivalent resistance is

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{6} + \dfrac{1}{6} = \dfrac{2}{6} = \dfrac{1}{3}$

Hence, R_p = 3Ω

The third resistor in series, then we get equivalent resistance

R_p + 6 = 3 + 6 = 9Ω

∴ To get a resistance of 9 Ω, two 6 Ω resistors should be connected in parallel and the third 6 Ω resistor should be connected in series with the combination as shown below:

Case (ii)

When two resistors are connected in series, their equivalent resistance is

R_s = 6 Ω + 6 Ω = 12 Ω

The third resistor is connected in parallel with 12 Ω. Hence the equivalent resistance is:

$\dfrac{1}{\text{R}} = \dfrac{1}{12} + \dfrac{1}{6} = \dfrac{1+2}{12} = \dfrac{3}{12} = \dfrac{1}{4}$

⇒ R = 4 Ω

∴ To get a resistance of 4 Ω, two 6 Ω resistors should be connected in series and the third 6 Ω resistor should be connected in parallel with the combination as shown below:

Given,

V = 220 V

P = 10 W

I_Max = 5 A

The resistance of the each bulb:

R = $\dfrac{\text{V}^2}{\text{P}} = \dfrac{220^2}{10} = 4840 Ω$

Assuming n bulbs are connected in parallel

$\dfrac{1}{\text{R}_{eq}} = \dfrac{1}{\text{R}} + \dfrac{1}{\text{R}} + ....n \\[1em] \dfrac{1}{\text{R}_{eq}} = \dfrac{1}{4840} + \dfrac{1}{4840} + ....n \\[1em] \dfrac{1}{\text{R}_{eq}} = \dfrac{\text{n}}{4840} \Omega \\[1em] \Rightarrow \text{R}_{eq} = \dfrac{4840}{\text{n}} \Omega$

Applying Ohm's Law:

V = IR

220 = 5 x $\dfrac{4840}{\text{n}}$

n = $\dfrac{4840 \times 5}{220}$ = 110 lamps

Hence, 110 lamps can be connected in parallel.

Case 1 — When coils are used separately

Applying Ohm's law:

$\text{V} = \text{IR} \\[1em] \therefore \text{I} = \dfrac{\text{V}}{\text{R}} \\[1em] \Rightarrow \text{I} = \dfrac{220}{24} \\[1em] \Rightarrow \text{I} = 9.166 \text{ A} \approx 9.2 \text{ A}$

Hence, 9.2 A of current flows through each resistor when they are used separately.

Case 2 — When coils are connected in series

R_s = 24 Ω + 24 Ω = 48 Ω

Applying Ohm's law:

$\text{V} = \text{IR} \\[1em] \therefore \text{I} = \dfrac{\text{V}}{\text{R}} \\[1em] \Rightarrow \text{I} = \dfrac{220}{48} \\[1em] \Rightarrow \text{I} = 4.58 \text{ A} \approx 4.6 \text{ A}$

Therefore, the current is 4.6 A when the two coils are used in series.

Case 3 — When coils are connected in parallel

$\dfrac{1}{\text{R}_p} = \dfrac{1}{24} + \dfrac{1}{24} = \dfrac{2}{24} = \dfrac{1}{12}$

∴ R_p = 12 Ω

Applying Ohm's law:

$\text{V} = \text{IR} \\[1em] \therefore \text{I} = \dfrac{\text{V}}{\text{R}} \\[1em] \Rightarrow \text{I} = \dfrac{220}{12} \\[1em] \Rightarrow \text{I} = 18.3 \text{ A}$

Therefore, the current is 18.3 A when the two coils are used in parallel.

(i) The potential difference is 6 V and the resistors 1 Ω and 2 Ω are connected in series:

R_s = 1 Ω + 2 Ω = 3 Ω.

Applying Ohm's law:

$\text{V} = \text{IR} \\[1em] \therefore \text{I} = \dfrac{\text{V}}{\text{R}} \\[1em] \Rightarrow \text{I} = \dfrac{6}{3} \\[1em] \Rightarrow \text{I} = 2 \text{ A}$

Power across 2 Ω resistor = I²R = 2 x 2 x 2 = 8 W

Hence, the power consumed by the 2 Ω resistor is 8 W.

(ii) In parallel circuit, voltage across the resistors remains the same. So, voltage across 2 Ω resistor is 4 V:

Power across 2 Ω resistor = $\dfrac{\text{V}^2}{{\text{R}}} = \dfrac{4 \times 4}{2}$ = 8 W

The power consumed by the 2 Ω resistor is 8 W.

Hence, same power is consumed by 2 Ω resistor in both the circuits.

As the two lamps are connected in parallel, the voltage across each of them will be the same.

Let I₁ be the current drawn by 100 W bulb:

I₁ = $\dfrac{\text{P}}{\text{V}} = \dfrac{100}{220} = \dfrac{5}{11} \text{ A}$

Let I₂ be the current drawn by 60 W bulb:

I₂ = $\dfrac{\text{P}}{\text{V}} = \dfrac{60}{220} = \dfrac{3}{11} \text{ A}$

∴ Current drawn from line = I₁ + I₂

$= \dfrac{5}{11} + \dfrac{3}{11} \\[1em] = \dfrac{5 + 3}{11} \\[1em] = \dfrac{8}{11} \text{ A} \\[1em] = 0.727272 \text{ A} \approx 0.73 \text{ A}$

Hence, 0.73 A current flows through the line.

Given,

Power of TV set = P₁ = 250 W

Time taken by TV set = t₁ = 1 hr = 3600 s

Power of toaster = P₂ = 1200

Time taken by toaster = t₂ = 10 minutes = 600 s

We know, energy consumed (H) = Pt

Energy consumed by TV set:

E₁ = 250 × 3600 = 900000 J = 9 × 10⁵ J

Energy consumed by toaster:

E₂ = 1200 × 600 = 720000 = 7.2 × 10⁵ J

As E₁ > E₂, hence, energy consumed by the TV set is greater than the toaster.

Given,

I = 15 A

R = 8 Ω

The rate at which the heat develops in the heater (P) = I²R

Substituting the values in the equation, we get

P = 15 x 15 x 8 = 1800 watt

Therefore, electric heater produces heat at the rate of 1800 W

(a) The resistivity and melting point of tungsten is very high hence it doesn't burn readily when heated. Therefore, tungsten is used almost exclusively for filament of electric lamps.

(b) Resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidize (burn) readily at high temperatures. For this reason, they are commonly used in electrical heating devices like electric iron, toasters etc.

(c) The series arrangement is not used for domestic circuits due to the following reasons:

1. All the connected appliances cannot be operated independently. If one device is defective, then the entire circuit will not function.
2. The overall voltage gets distributed in a series circuit. As a result, electric appliances may not get the rated power for their operation.
3. The total resistance becomes large, hence, the current is reduced.

(d) Resistance (R) of a wire is inversely proportional to its area of cross-section (A) i.e., when the area of cross section increases the resistance decreases and vice versa.
R = ρ $\dfrac{l}{A}$

(e) Copper and aluminium both are good conductors of electricity and have low resistivity. Hence, they can efficiently carry electrical current with minimal resistance, which reduces energy losses as heat during transmission.

A continuous and closed path of an electric current is called an electric circuit.

The unit of current is ampere (A). One ampere is constituted by the flow of one coulomb of charge per second, i.e., 1A = $\dfrac{1 \text{C}}{1 \text{s}}$

Charge on one electron (e) = 1.6 × 10^-19 C

q = 1 C

Let n electrons constitute one coulomb of charge.

∴ q = ne

where,

q = charge
e = charge on one electron
n = number of electrons

⇒ n = $\dfrac{\text{q}}{\text{e}}$

⇒ n = $\dfrac{1}{1.6 \times 10^{-19}}$

= 6.25 x 10^-18 ≈ 6 x 10^-18

Therefore, the number of electrons constituting one coulomb of charge is 6 x 10^-18

Battery/Cell.

When 1 Joule of work is done to move a charge of 1 coulomb from one point to another, then the potential difference between the two points is said to be 1 V.

$\therefore 1 \text{ Volt} = \dfrac{1 \text{ Joule}}{1 \text{ Coulomb}}$

Given,

Potential difference (V) = 6V

Charge (q) = 1 C

We know,

V = $\dfrac{W}{q}$

Hence, Work done for 1 C will be W = Vq

Substituting we get, W = 6 x 1 = 6 J

Hence, 6 J of energy is given to each coulomb of charge passing through a 6 V battery.

The resistance of a conductor depends on :

1. Its length
2. Its area of cross section
3. The nature of its material

R = ρ $\dfrac{l}{A}$

Area of the cross-section of wire is inversely proportional to the resistance, so, the thick wire offers less resistance, hence, current will flow more easily through it.

According to Ohm's Law:

I = $\dfrac{\text{V}}{\text{R}}$

Given, the potential difference decreases to half of its former value, keeping the resistance constant

V₂ = $\dfrac{\text{V}}{2}$ , R₂ = R, I₂ = ?

Substituting the values we get,

I₂ = $\dfrac{\text{V}}{2\text{R}}$ = $\dfrac{\text{I}_1}{2}$

Therefore, the current flowing through the electrical component is reduced by half.

Resistivity of an alloy is generally higher than that of its constituent metals. Alloys do not oxidize (burn) readily at high temperatures. For this reason, they are commonly used in electrical heating devices like electric iron, toasters etc.

(a) Iron with a lesser resistivity of 10.0 x 10^-8 is a better conductor than mercury having a resistivity of 94.0 x 10^-8.

(b) Silver is the best conductor because it's resistivity of 1.60 x 10^-8 is the least in the table.

According to Ohm's Law,

V = IR

The total resistance of the circuit is 5 Ω + 8 Ω + 12 Ω = 25 Ω.

Potential difference (V) = 6 V,

I = $\dfrac{\text{V}}{\text{R}} = \dfrac{6}{25}$ = 0.24 A

Hence, current shown in ammeter = 0.24 A

Let potential difference across the 12 Ω resistor be V₁.

V₁ = 0.24 × 12 = 2.88 V

Therefore, the voltmeter reading will be 2.88 V.

(a) 1 Ω and 10⁶ Ω are connected in parallel, their equivalent resistance will be:

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{\text{1}} + \dfrac{1}{10^6} = \dfrac{10^6 +1}{10^6}$

R_p = $\dfrac{10^6}{10^6 +1}$ ≈ $\dfrac{10^6}{10^6}$

R_p = 1 Ω

Therefore, the equivalent resistance is 1 Ω (approx).

(b) 1 Ω, 10³ Ω, and 10⁶ Ω are connected in parallel, their equivalent resistance will be:

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{\text{1}} + \dfrac{1}{10^3} + \dfrac{1}{10^6} = \dfrac{10^6 + 10^3 + 1}{10^6} = \dfrac{1001001}{1000000}$

R_p = $\dfrac{1000000}{1001001}$ = 0.999 Ω

Therefore, the equivalent resistance is 0.999 Ω.

Given,

Potential difference across circuit = 220 V

Resistance of electric lamp = 100 Ω

Resistance of toaster = 50 Ω

Resistance of water filter = 500 Ω

Equivalent resistance across the three appliances connected in parallel

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{\text{100}} + \dfrac{1}{\text{50}}+ \dfrac{1}{\text{500}} = \dfrac{5+10+1}{\text{500}} = \dfrac{16}{\text{500}}$

Hence, R_p = $\dfrac{500}{\text{16}} = \dfrac{125}{\text{4}} = 31.25 \space \Omega$

The resistance of the electric iron connected to the same source will be 31.25 Ω as same current flows through it as that across the three appliances (connected in parallel).

Current across electric iron = ?

Applying Ohm's law

V = IR

Substituting we get,

220 = I x 31.25

I = $\dfrac{220}{31.25}$ = 7.04 A

Therefore, current through the electric iron will be 7.04 A.

Advantages of parallel connection over series connection are:

1. Independent Operation — Devices connected in parallel operate independently of each other. Failure of one device does not affect the others. In series connection, if one device fails, the whole circuit goes down.
2. Voltage Stability — In parallel connection, each device receives the same voltage as the battery regardless of the number of devices connected. This ensures consistent device performance and prevents the voltage drop that occurs in series circuits as more devices are added.
3. Reduced Total Resistance — In parallel connection, the effective resistance of the circuit is reduced. This is helpful when each gadget has a different resistance and requires different current to operate properly.
4. Easy Troubleshooting — Troubleshooting parallel circuits is often easier because a fault in one device does not affect the others. In a series circuit, locating a fault can be more challenging because it disrupts the entire circuit.

(a) If we connect 3 Ω and 6 Ω in parallel, the equivalent resistance will be less than 3 Ω. Now, if 2 Ω is connected in series with this equivalent resistance, we should get a total resistance of 4 Ω.

So, the circuit will be as shown below:

3 Ω and 6 Ω are connected in parallel

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{2+1}{6} = \dfrac{3}{6} = \dfrac{1}{2}$Ω

Hence, R_p = 2 Ω

The equivalent resistor 2 Ω is in series with the 2 Ω resistor.

R_eq= 2 Ω + 2 Ω = 4 Ω

Hence, the total resistance of the circuit is 4 Ω.

(b) Since the total resistance in this case is less than the resistance of each resistor hence, the resistors 2 Ω, 3 Ω and 6 Ω must be connected in parallel to get a total resistance of 1 Ω.

So, the circuit will be as shown below:

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{3+2+1}{6} = \dfrac{6}{6} = 1Ω$

R_eq = 1Ω

Hence, the total resistance of the circuit is 1 Ω.

(a) To secure the highest total resistance, the four coils should be connected in series as shown below:

R_eq = 4 Ω + 8 Ω + 12 Ω + 24 Ω = 48 Ω.

(b) To secure the lowest total resistance, the four coils should be connected in parallel as shown below:

$\dfrac{1}{\text{R}_\text{p}} = \dfrac{1}{4} + \dfrac{1}{8} + \dfrac{1}{12} + \dfrac{1}{24} \\[1em] = \dfrac{6+3+2+1}{24} \\[1em] = \dfrac{12}{24} = \dfrac{1}{2} \Omega$

R_eq = 2Ω

Hence, the lowest total resistance is 2 Ω.

The heating element of an electric heater is made of an alloy having high resistance and when the current flows through it, the heating element becomes too hot and glows red. The cord on the other hand is made of copper or aluminum having low resistance, so current easily flows through it producing minimal heat. Hence, it doesn't glow.

Given,

t = 1 hour = 3600 s
V = 50 V
Q = 96000 C

I = $\dfrac{\text{Q}}{\text{t}} = \dfrac{96000}{3600} = \dfrac{80}{3}$A

According to Joule's law :

Heat (H) = Voltage (V) x current (I) x time (t)

Substituting we get,

H = 50 x $\dfrac{80}{3}$ x 3600 = 4.8 x 10⁶ J

Therefore, amount of heat generated = 4.8 x 10⁶ J

Given,

R = 20 Ω
I = 5 A
t = 30 s

According to Joule's law of heating,

H = I²Rt

Substituting the values, we get,

H = 5 x 5 x 20 x 30 = 15000 J

Therefore, amount of heat developed = 15000 J.

Power determines the rate at which energy is delivered by a current.

Given,

I = 5 A
V = 220 V
t = 2 hr = 7200 s
P = ?

We know

Power (P) = Voltage (V) x Current (I)

Substituting the values, we get

P = 220 × 5 = 1100 W

The energy consumed by the motor (E) = P × T

Substituting the values, we get

E = 1100 × 7200

= 79,20,000 J

= 7.92 × 10⁶ J

Hence, power of the motor is 1100 W and the energy consumed is 7.92 × 10⁶ J.